Test Index

CBSE Class 12 Physics 2017 Outside Delhi Set 1 Paper

© examsnet.com
Question : 25 of 26
Marks: +1, -0
A device ' XX ' is connected to an ac source V=V0sin⁡ωtV=V_0 \sin \omega t. The variation of voltage, current and power in one cycle is shown in the following graph:
(a) Identify the device ' XX '.
(b) Which of the curves A, B and C represent the voltage, current and the power consumed in the circuit? Justify your answer.
(c) How does its impedance vary with frequency of the ac source? Show graphically.
(d) Obtain an expression for the current in the circuit and its phase relation with ac voltage.
OR
(a) Draw a labelled diagram of an ac generator. Obtain the expression for the emf induced in the rotating coil of NN turns each of cross-sectional area AA, in the presence of a magnetic field B→\overset{\rightarrow}{B}.
(b) A horizontal conducting rod 10 m10\text{ m} long extending from east to west is falling with a speed 5⋅0ms−15 \cdot 0 \text{ms}^{-1} at right angles to the horizontal component of the Earth's magnetic field, 0.3×10−4Wb/m20.3 \times 10^{-4} \mathrm{Wb/m^2}.
Find the instantaneous value of the emf induced in the rod.
Solution:  
(a) Identification
(b) Identifying the curves Justification
(c) Variation of Impedance with frequency Graph
(d) Expression for current Phase relation
(a) The device XX is a capacitor
(b) Curve B →\rightarrow voltage
Curve C →\rightarrow current
Curve A→A \rightarrow power
Reason: The current leads the voltage in phase,by ≠/2\neq / 2 for a capacitor.
(c) Xc=1ωc(Xc α1ω)X_c = \frac{1}{\omega c} \left(X_c \,\alpha \frac{1}{\omega}\right)
V=Vosin⁡ωtV=V_o \sin \omega t
q=CV=CVosin⁡ωtq = C V = C V_o \sin \omega t
I=dqdt=ωcVocos⁡ωtI = \frac{dq}{dt} = \omega c V_o \cos \omega t
=Iosin⁡(ωt+π2)= I_o \sin \left(\omega t + \frac{\pi}{2}\right)
Current leads the voltage, in phase, by ≠/2\neq / 2
OR
(a)
When thecoil rotatesin a magnetic field, itseffective area, i.e., Acos⁡θA \cos \theta, (i.e., area normal to the magnetic field) keeps on changing. Hence magnetic flux ϕ\phi =NBAcos⁡θ= N B A \cos \theta , keeps on changing. Let the coil be rotating with angular velocity ' ω\omega ', at any instant ' tt ' when the normal to the plane of the coil makes an angle θ\theta with the magnetic field. Hence magnetic flux,
ϕ=NBAcos⁡ωt,\phi = N B A \cos \omega t,
Therefore induced emf ϵ=−dϕdt\epsilon = -\frac{d\phi}{dt}
⇒ϵ=NBAωsin⁡ωt\Rightarrow \epsilon = N B A \omega \sin \omega t
induced emf will be maximum when ωt=90∘\omega t = 90^{\circ}
Hence, ϵmax⁡=NBAω\epsilon_{\max} = N B A \omega
The direction of induced emf can be determined using Fleming's right hand rule.
(b) I=10 m,u=5 m/s,\text{(b) } I=10\text{ m}, u=5\text{ m}/\text{s}, B=0.3×10−4Wb/m2B = 0.3 \times 10^{-4} \mathrm{Wb/m^2}
∴ϵ=Blv\therefore \epsilon = \text{Blv}
=0.3×10−4×10×5= 0.3 \times 10^{-4} \times 10 \times 5
=15×10−4V= 15 \times 10^{-4} \mathrm{V}
  \;
© examsnet.com
Go to Question:
Prev Question
Next Question